Float drift and output

Questions 1 to 5 of 6

1. 1. What does this program print? ``cpp #include <iostream> int main() { double sum = 0.0; for (int i = 0; i < 10; ++i) { sum += 0.1; } std::cout << sum << ' ' << (sum == 1.0) << '\n'; } ``

   1 10 00.99999999999999989 01 0
2. 2. What does this program print? ``cpp #include <iostream> int main() { int count = 0; for (double t = 20.0; t <= 21.0; t += 0.1) { ++count; } std::cout << "count: " << count << '\n'; } ``

   count: 9count: 10count: 11count: 21
3. 3. True or false, with a reason: for (double x = 0.0; x <= 1.0; x += 0.25) prints exactly five values, 0.0, 0.25, 0.5, 0.75 and 1.0.

   False, because 0.25 cannot be stored exactly, so the last value misses 1.0True, because 0.25 is a sum of powers of one half, so every value is exact and 1.0 is reached
4. 4. True or false, with a reason: printing a double with the default precision shows the drift, so ten additions of 0.1 print as 0.99999999999999989.

   False, because the default precision prints about six significant digits, so the sum shows as 1True, because the default precision prints every stored digit of the sum
5. 5. Why must the counter loop divide by 10.0 in double t = 20.0 + i / 10.0; and not by 10?

   Because 10.0 makes the program faster to compileBecause the compiler rejects i / 10 with an errorBecause with integers the division drops the fraction, so i / 10 is 0 for every i below 10Because i / 10 stores a double with more drift